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  • Explain Solution R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Function Exercise Multiple Choice Questions Question 44 maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Function Exercise Multiple Choice Questions Question 44 maths Textbook Solution.

Answers (1)

Answer: \sqrt{5}-2

Hint: \tan \theta=\frac{\sin \theta}{\cos \theta}

Given: \tan \left(\frac{1}{2} \cos ^{-1} \frac{2}{\sqrt{5}}\right)

Solution:

Let, \left(\frac{1}{2} \cos ^{-1} \frac{2}{\sqrt{5}}\right)=\theta

\begin{aligned} &\cos ^{-1} \frac{2}{\sqrt{5}}=2 \theta \\ &\frac{2}{\sqrt{5}}=\cos 2 \theta \end{aligned}

\begin{aligned} &2 \cos ^{2} \theta-1=\frac{2}{\sqrt{5}} \\ &2 \cos ^{2} \theta=\frac{1}{\sqrt{5}} \\ &\cos \theta=\sqrt{\frac{1}{2}+\frac{1}{\sqrt{5}}} \end{aligned}

\tan \theta=\frac{\sin \theta}{\cos \theta}

           \begin{aligned} &\frac{\sqrt{\frac{1}{2}-\frac{1}{\sqrt{5}}}}{\sqrt{\frac{1}{2}+\frac{1}{\sqrt{5}}}} \\ &=\sqrt{\frac{(\sqrt{5}-2)^{2}}{(\sqrt{5}-2)(\sqrt{5}+2)}} \end{aligned}

             =(\sqrt{5}-2) Alignment needs improvement

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